Question

A plane meets the coordinate axes at points A, B, C and $(\alpha ,\beta ,\gamma )$ is the centroid of the triangle ABC. Then the equation of the plane is

• A. $\frac{x}{\alpha }+\frac{y}{\beta }+\frac{z}{\gamma }=3$
• B. $\frac{x}{\alpha }+\frac{y}{\beta }+\frac{z}{\gamma }=1$
• C. $\frac{3x}{\alpha }+\frac{3y}{\beta }+\frac{3z}{\gamma }=1$
• D. $\alpha x+\beta y+\gamma z=1$

Answer 1

The correct option is A

$\frac{x}{\alpha }+\frac{y}{\beta }+\frac{z}{\gamma }=3$

Let the co-ordinates of the points where the plane cuts the axes be (a,0,0), (0,b,0) and (0,0,c).
Since the centroid is $(\alpha ,\beta ,\gamma )$, we have $a=3\alpha ,b=3\beta ,c=3\gamma$
The equation of the plane will be
$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\Rightarrow \frac{x}{3\alpha }+\frac{y}{3\beta }+\frac{z}{3\gamma }=1\Rightarrow \frac{x}{\alpha }+\frac{y}{\beta }+\frac{z}{\gamma }=3$